Question

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 
Values val[16];

How can one allocate memory for numbers in the struct?

Answer 1

So, you could do this without using dynamic memory allocation like this:

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 

#define MAX_SIZE 16

int main() {
    Values val[MAX_SIZE];
    float myfloats[MAX_SIZE];
    int i;
    for(i=0;i<MAX_SIZE;i++) {
        val[i].numbers=&myfloats[i];
    }
    return 0;
 }

But I can’t think of any reason why you’d want a structure with a pointer to just one float.

Based on the name 'numbers', I’d say you want 'numbers' to point to an array of floats, if so, you could do this:

#include <malloc.h>

typedef struct {
    float *numbers;
    float val1;
    float val2;
} Values; 

#define MAX_SIZE 16

int main() {
    Values val[MAX_SIZE];
    size_t numberOfFloats = 10;
    int i;

    // for each of the members of the val array
    for(i=0;i<MAX_SIZE;i++) {
        // allocate using calloc (this will set all of the floats to 0.0)
        val[i].numbers=calloc(numberOfFloats,sizeof(float));

        // check the allocation worked...
        if(!val[i].numbers) {
            // insert proper error handling here.
            printf("oops\n");
            return -1;
        }

    }

    // you access the variables like this
    for(i=0;i<MAX_SIZE;i++) {
        int number;
        for(number=0; number < numberOfFloats; number++) {
            printf("Value %d, Number %d = %f\n",i,number,val[i].numbers[number]);
        }
    }

    // don't forget to play nice and clean up afterwards
    for(i=0;i<MAX_SIZE;i++) {
        free(val[i].numbers);
    }
    return 0;
 }